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我不太确定为什么以下代码在 GCC 4.6.3 中给我以下错误

'boost::spirit::_a = boost::phoenix::function::operator()(const A0&) const [with A0 = boost::phoenix::actor >, F = make_line_impl 中的 'operator=' 不匹配, 类型名 boost::phoenix::as_composite, F, A0>::type = boost::phoenix::composite, boost::fusion::vector, boost::spirit::argument<0>, boost::fusion: :void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_, boost::fusion::void_ , boost::fusion::void_>]((* & boost::spirit::_1))'</p>

甚至可以将惰性函数对象的结果分配给 qi 占位符吗？

#include <string>
#include <boost/spirit/include/phoenix_function.hpp>
#include <boost/spirit/include/qi.hpp>

using std::string;

using boost::spirit::qi::grammar;
using boost::spirit::qi::rule;
using boost::spirit::qi::space_type;
using boost::spirit::qi::skip_flag;
using boost::spirit::unused_type;

namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;

struct make_line_impl
{
  int* _context;
  make_line_impl(int* context)
  {
    _context = context;
  }

  template <typename Sig>
  struct result;

  template <typename This, typename Arg>
  struct result<This(Arg const &)>
  {
    typedef int* type;
  };
  template <typename Arg>
  int* operator()(Arg const & content)
  {
    return new int(5);
  }
};


template<typename Iterator>
struct MyGrammar : grammar<Iterator, unused_type, space_type>
{
  rule<Iterator, unused_type, space_type> start;
  rule<Iterator, int*(), space_type> label;
  rule<Iterator, string*(), qi::locals<int*>, space_type> line;

  MyGrammar() : MyGrammar::base_type(start)
  {
    make_line_impl mlei(new int(5));
    phx::function<make_line_impl> make_line(mlei);

    start = *(line);

    line = label[qi::_a = make_line(qi::_1)];
  }
};


int main(int argc, char **argv) {

  string contents;

  qi::phrase_parse(contents.begin(), contents.end(), MyGrammar<string::iterator>(), space_type(), skip_flag::postskip);
    return 0;
}
4

1 回答 1

2

我已经修复了代码中的一些问题以使其编译：

  • ::result<>::type我根据BOOST_RESULT_OF的文档重写了嵌套逻辑。

    注意如果你在 c++11 模式下编译，你最好定义

     #define BOOST_RESULT_OF_USE_DECLTYPE

    在这种情况下，您不必费心使用嵌套结果类型模板。

  • operator(...)方法需要是const

结果代码：

#include <string>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/qi.hpp>

using std::string;

using boost::spirit::qi::grammar;
using boost::spirit::qi::rule;
using boost::spirit::qi::space_type;
using boost::spirit::qi::skip_flag;
using boost::spirit::unused_type;

namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;

struct make_line_impl
{
  int* _context;
  make_line_impl(int* context)
  {
    _context = context;
  }

  template <typename Arg> struct result { typedef int* type; };

  template <typename Arg>
  int* operator()(Arg const & content) const
  {
    return new int(5);
  }
};


template<typename Iterator>
struct MyGrammar : grammar<Iterator, unused_type, space_type>
{
  rule<Iterator, unused_type, space_type> start;
  rule<Iterator, int*(), space_type> label;
  rule<Iterator, string*(), qi::locals<int*>, space_type> line;

  MyGrammar() : MyGrammar::base_type(start)
  {
    make_line_impl mlei(new int(5));
    phx::function<make_line_impl> make_line(mlei);

    start = *(line);

    line = label[qi::_a = make_line(qi::_1)];
  }
};


int main(int argc, char **argv) {

  string contents;

  qi::phrase_parse(contents.begin(), contents.end(), MyGrammar<string::iterator>(), space_type(), skip_flag::postskip);
    return 0;
}
于 2012-05-04T08:27:28.077 回答